Reactions & Stoichiometry

Report
UNIT 6
Overview

Reactions






Write formula/word equations
Balance Equations
Identify Types
Predict Products
Write Net Ionic Equations
Stoichiometry
 Conversions
 Limiting & Excess Reagent
 Percent Yield
Chemical Reactions

Process in which one or more pure substances are
converted into one or more different pure substances
Reactants: Zn + I2
Product: Zn I2
Indications of a Reaction
Temperature Change
 Color Change
 Production of gas
 Formation of a precipitate
 Production of light

Chemical Equations
4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
(Reactants)
(Products)

Reactants react to produce products

The letters (s), (g), (l), and (aq) are the physical
states of compounds.
 “aq” represents aqueous meaning dissolved in water (solution)

The numbers in the front are called coefficients.

Subscripts represent the number of each atom in a
compound
Chemical Reactions
Symbol
Meaning
+
→
used to separate the reactants from the products - it is
pronounced "yields" or "produces" when the equation is read
↔
used when the reaction can proceed in both directions - this is
called an equilibrium arrow and will be used later in the course
↑
↓
∆
used to separate one reactant or product from another
an alternative way of representing a substance in a gaseous
state
an alternative way of representing a substance in a solid state
indicates that heat is applied to make the reaction proceed
Diatomic Elements

Elements that cannot exist by
themselves (always occur in pairs)
 Bromine (Br2)
 Iodine (I2)
 Nitrogen (N2)
 Chlorine (Cl2)
 Hydrogen (H2)
 Oxygen (O2)
 Fluorine (F2)
Writing Equations Practice
1. When lithium hydroxide pellets are added to a
solution of sulfuric acid, lithium sulfate and water
are formed.
2 LiOH(s) + H2SO4(aq)  Li2SO4(aq) + 2 H2O(l)
2. When crystalline C6H12O6 is burned in
oxygen, carbon dioxide and water vapor are
formed.
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
Balancing Equations

Law of Conservation of Mass
 Matter cannot be destroyed (atoms of
reactants must equal products)

Balance equations to get same number of
each atom on the left and right in an
equation
2HgO(s) ---> 2 Hg(l) + O2(g)
2 Hg atoms, 2 O atoms
2 Hg atoms, 2 O atoms
Balancing Equations
___ Al(s) + ___ Br2(l) ---> ___ Al2Br6(s)
__C3H8(g) + __ O2(g)  __ CO2(g) + __ H2O(g)
__ B4H10(g) + __ O2(g)  __ B2O3(g) + __ H2O(g)
6 Types of Reactions
Synthesis (combination)
 Decomposition
 Single Replacement (displacement)
 Double Replacement (precipitation)
 Combustion
 Acid-Base Neutralization

Synthesis (Combination) Reactions

Two or more substances combine to form a
new compound.
A + X  AX
Synthesis of:
Binary compounds
Metal carbonates
Metal hydroxides
Metal chlorates
Oxyacids
H2 + O2  H2O
CaO + CO2  CaCO3
CaO + H2O  Ca(OH)2
KCl + O2  KClO3
CO2 + H2O  H2CO3
Decomposition Reactions

A single compound breaks down into two
or more simpler substances
AX  A + X
Decomposition of:
Binary compounds
Metal carbonates
Metal hydroxides
Metal chlorates
Oxyacids
H2O  H2 + O2
CaCO3  CaO + CO2
Ca(OH)2  CaO + H2O
KClO3  KCl + O2
H2CO3  CO2 + H2O
Single Replacement
(displacement) Reactions

One element replaces another in a
reaction
 Metals replace metals
 Nonmetals replace nonmetals
A + BX  AX + B
BX + Y  BY + X
Activity Series
Decide whether or not one element will
replace another
 Metals can replace other metals provided
that they are above the metal that they are
trying to replace
 If the metal is not
above what it is
trying to replace,
the result is “no
reaction”

Double Replacement
(Precipitation) Reactions

Two elements or ions “switch partners”
AX + BY  AY + BX

One of the compounds
formed is usually a
precipitate, an insoluble
gas that bubbles out of
solution, or a molecular
compound, usually water.
Solubility


Solubility – ability to dissolve
In a double replacement (precipitate) reaction,
one of the products must be insoluble in water
and form a precipitate
 Precipitate
– insoluble solid formed by a reaction in
solution
 If both products are soluble the result is “no reaction”

Solubility rules help you determine whether or not
a compound will form a precipitate or remain an
aqueous solution
Solubility Rules
Soluble Ionic Compounds
Except with:
Alkali metals, NH4+
NO3-, C2H3O2-, ClO3-, ClO4-
(no exceptions)
Cl-, Br-, I-
Ag+, Hg2+2, Pb+2
SO4-2
Sr+2, Ba+2, Ca+2, Ag+, Pb+2, Hg2+2
Insoluble Ionic Compounds
Except with:
CO3-2, PO4-3, SiO3-2, O-2, SO3-2, CrO4-2
NH4+, alkali metals
S-2
NH4+, alkali metals
Ca+2, Sr+2, Ba+2, Mg+2 (group 2)
OH-
NH4+, alkali metals,
(Ca+2, Ba+2, Sr+2 are slightly soluble)
Combustion Reactions

A substance combines with oxygen,
releasing a large amount of energy in the
form of light and heat.
 Produces a flame
 Fuel + oxygen produces carbon dioxide and
water vapor
CxHx + O2  CO2 + H2O
Acid-Base Neutralization Reactions
When the solution of an acid and solution of a
base are mixed
 Products have no characteristics of either the
acid or the base
 Acid + Base (metal hydroxide)  salt + water

 Salt comes from cation of base and anion of acid
HY + XOH  XY + H2O
Chemical Equations

Molecular Equation – shows complete chemical
formulas of reactants and products
Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)

Complete Ionic Equation – All soluble electrolytes
shown as ions
Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq)

Net Ionic Equation – shows only the ions and molecules
directly involved in the equation
Pb+2(aq) + 2I-(aq)  PbI2(s)
Writing Complete Ionic Equations
Start with a balanced molecular equation.
2. Break all soluble strong electrolytes
(compounds with (aq) beside them) into
their ions.
1.



3.
indicate the correct formula and charge of
each ion
indicate the correct number of each ion
write (aq) after each ion
Bring down all compounds with (s), (l), or
(g) unchanged.
Writing Complete Ionic Equations
Example:
2Na3PO4(aq) + 3CaCl2(aq)  6NaCl(aq) + Ca3(PO4)2(s)
Becomes…
6Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq)  6Na+(aq) +
6Cl-(aq) + Ca3(PO4)2(s)
Spectator Ions

Appear in identical forms among both
the reactants and products of a
complete ionic equation

When writing net ionic equations they
cancel each other out
Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq)
Writing Net Ionic Equations

Cancel out spectator ions from complete ionic
equation then write what’s left
6Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq)  6Na+(aq) +
6Cl-(aq) + Ca3(PO4)2(s)
Becomes…
2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2(s)
Practice
Write complete ionic and net ionic equations for
the following:
1.
3(NH4)2CO3(aq) + 2Al(NO3)3(aq)  6NH4NO3(aq) +
Al2(CO3)3(s)
2.
2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(l)
3.
Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s)
Answers
1.
Complete Ionic Equation:
6NH4+(aq) + 3CO32-(aq) + 2Al3+(aq) + 6NO3-(aq)  6NH4+(aq) + 6NO3-(aq) + Al2(CO3)3(s)
Net Ionic Equation:
2.
2 Al3+(aq) + 3 CO32-(aq)  Al2(CO3)3(s)
Complete Ionic Equation:
2Na+(aq) + 2OH-(aq) + 2H+(aq) + SO42-(aq)  2Na+(aq) + SO42-(aq) + 2H2O(l)
Net Ionic Equation:
OH-(aq) + H+(aq)  H2O(l)
*Note: simplify net ionic equations if possible
3.
Complete Ionic Equation:
Zn(s) + Cu2+(aq) + SO42-(aq)  Zn2+(aq) + SO42-(aq) + Cu(s)
Net Ionic Equation:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Stoichiometry
The study of the
quantitative
aspects of
chemical
reactions.
Mole Ratio

Conversion factor that relates amount in
moles of any two substances involved in
a chemical reaction
2 Al2O3(l)  4 Al(s) + 3 O2(g)
Mole ratio Al2O3 to O2 = 2:3
Mole ratio Al to Al2O3 = 4:2 or 2:1
Mole ratio Al to O2 = 4:3
Stoichiometry Problems
Solved just like conversions!
 You must start with a balanced chemical
equation
 Types:

 Mole  Mole
 Mass  Mass
 Mass  Mole or Mole  Mass
Mole  Mole
2 Al2O3(l)  4 Al(s) + 3 O2(g)

How many moles of O2 are produced from
3.5 moles of Al2O3?
3 mol O2
3.5 mol Al2O3 ×
= 5.25 mol O2
2 mol Al2O3
*Use mole ratio to convert between moles!
Mass  Mass
2 Al2O3(l)  4 Al(s) + 3 O2(g)

How many grams of Al are produced from
4.56 grams of Al2O3?

Molar Mass Al2O3 = 101.96 g/mol
4.56 g Al2O3 ×
Molar Mass Al = 26.98 g/mol
1 mol Al2O3
26.98 g Al
4 mol Al
×
×
= 2.41 g Al
101.96 g Al2O3 2 mol Al2O3
1 mol Al
Limiting/Excess Reactant
Recipe makes 10 pancakes




3 eggs
2 cups bisquik
1 cup milk
1 cup chocolate chips
What “limits” how
many pancakes I
can make and what
will be left over?

What is the most amount of pancakes that I can
make with 6 eggs and 5 cups of milk?

What is the most amount of pancakes that I can
make with 3 cups of chocolate chips and 8 cups of
milk?
Limiting/Excess Reactant
The limiting reactant is the reactant that is
consumed first, limiting the amounts of
products formed.
 The excess reactant is the reactant that is
leftover after the reaction has gone to
completion.

Limiting/Excess Reactant
Reactants
2 NO(g) + O2 (g)
Products
2 NO2(g)
Limiting reactant = ___________
Excess reactant = ____________
Calculating Limiting/Excess Reagent
2 NO(g) + O2 (g)

2 NO2(g)
Given 12.4 grams of NO and 9.40 grams
of O2, which is the limiting and which is
the excess reagent?
12.4 g NO ×
1 mol NO
46.01 g NO2
2 mol NO2
×
×
= 19.01 g NO2
30.01 g NO
1
mol
NO
2 mol NO
2
9.40 g O2 ×
1 mol O2 × 2 mol NO2 × 46.01 g NO2 = 54.06 g NO
2
16.00 g O2
1 mol NO2
1 mol O2
Calculating Limiting/Excess Reagent
2 NO(g) + O2 (g)
2 NO2(g)
12.4 g NO ×
1 mol NO
46.01 g NO2
2 mol NO2
×
×
= 19.01 g NO2
30.01 g NO
1 mol NO2
2 mol NO
9.40 g O2 ×
1 mol O2 × 2 mol NO2 × 46.01 g NO2 = 27.03 g NO
2
32.00 g O2
1 mol NO2
1 mol O2

NO limits the amount of NO2 that is made
 Limiting reagent = NO

O2 will be leftover once the reaction is complete
 Excess reactant = O2
Calculating Limiting/Excess Reagent
2 NO(g) + O2 (g)

2 NO2(g)
How much O2 will be in excess once the
reaction is complete?
1 mol NO
32.00 g O2
1 mol O2
12.4 g NO × 30.01 g NO × 2 mol NO × 1 mol O = 6.61 g O2
2
6.61 grams of O2 will be used in the reaction.
You have 9.40 grams to start with.
9.40 – 6.61 = 2.79 grams O2 in excess (leftover)
Limiting/Excess Reactant

If the equation has 2 or more products,
when determining the limiting/excess
reactants, simply pick one of the products
and convert both reactants to that product.
 You MUST use the same product for both.
Percent Yield
Actual Yield
Theoretical Yield
× 100
Percentage comparing how much product
was actually produced compared to what
should have been produced.
 Calculate theoretical yield using
stoichiometry.

 If you know how much of each reactant you start
out with, use stoichiometry to calculate how much
of the given product you should produce.
Percent Yield
AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)
An experiment was performed combining using 3.4 g of
AgNO3 and an unlimited supply of KCl. If the
experiment yielded 2.7 g of AgCl, what is the percent
yield of the experiment?
3.4 g AgNO3 × 1 mol AgNO3 × 1 mol AgCl × 143.32 g AgCl = 2.9 g AgCl
169.88 g AgNO3 1 mol AgNO3
1 mol AgCl
2.7
Percent Yield = 2.9 × 100 = 93%

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